If one of the lines represented by \(a x^2+2 h x y+b y^2=0\) is perpendicular to \(\mathrm{m} x+\mathrm{n} y=18\), then

Given equation of pair of lines is \(\mathrm{a} x^2+2 \mathrm{~h} x y+\mathrm{b} y^2=0\)
\(\begin{aligned}
& \Rightarrow \mathrm{a}+2 \mathrm{~h}\left(\frac{y}{x}\right)+\mathrm{b}\left(\frac{y}{x}\right)^2=0 \\
& \Rightarrow \mathrm{a}+2 \mathrm{hk}+\mathrm{bk}^2=0\quad \ldots (i)
\end{aligned}\)
Now, slope of line \(m x+n y=18\) is \(\frac{-m}{n}\)
\(\therefore \) Slope of the line perpendicular to \(\mathrm{m} x+\mathrm{n} y=18\) is \(\mathrm{k}=\frac{\mathrm{n}}{\mathrm{m}}\)
Substituting the value of \(k\) in (i), we get
\(\begin{aligned}
& a+2 h\left(\frac{n}{m}\right)+b\left(\frac{n}{m}\right)^2=0 \\
& \Rightarrow a^2+2 h m n+n^2=0
\end{aligned}\)